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C++ tilida takrorlanuvchi jarayonlar dasturlar tuzish. For,while va do while takrorlash operatorlari. lab

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08.08.2023

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Labaratoriya №2
Mavzu: C++ tilida takrorlanuvchi jarayonlar dasturlar tuzish. For,while va do while
takrorlash operatorlari.
Misollar  1. Misol. (Namuna)  Butun son berilgan agar a>0 bo’lsa a+1 aks holsa a 
hisoblansin.
#include <iostream.h>
#include <math.h>
int main ()
 { int a;
 cin>>a;
 if(a>0){a++;}
 cout<<a;
 return 0;
 }
Javob:
1-misol.
Dastur kodi:
#include <iostream>
using namespace std;
int main()
{
   int n;
   double s=0;
   cin>>n;
   for(int i=1; i<=n; i++)
    {
        s+=1./i;
    }
   cout<<s;
   return 0;
} Natija:
2-misol.
Dastur kodi:
#include <iostream>
using namespace std;
int main()
{
   int n, s=0;
   cin>>n;
   for(int i=n; i<=2*n; i++)
    {
        s+=i*i;
    }
   cout<<s;
   return 0;
} Natija:
3-misol.
Dastur kodi:
#include <iostream>
using namespace std;
int main()
{
   int n;
   double  p=1,k=1;
   cin>>n;
   for(int i=1; i<=n; i++)
    {   k+=0.1;
        p*=k;
    }
   cout<<p;
   return 0;
} Natija :
4-misol.
Dastur kodi:
#include <iostream>
using namespace std;
int main()
{
   double A, p;
   int n;
   cin>>A;
   cin>>n;
   p=A;
   for(int i=1; i<=n; i++)
   {
       cout<<p<<" ";
       p*=A;
   }
   return 0;
} Natija :
5-misol.
Dastur kodi:
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
   double A, p, s=1;
   int n;
   cin>>A;
   cin>>n;
   p=A;
   for(int i=1; i<=n; i++)
   {
       s+=p;
       cout<<p<<" ";
       p*=A;
   }
   cout<<"\n"<<s;
   return 0;
} Natija :
6-misol.
Dastur kodi:
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
   double x, p, s=1;
   int n;
   cin>>x;
   cin>>n;
   p=x;
   for(int i=1; i<=n; i++)
   {
       s+=pow((-1),i)*p;
       cout<<p<<" ";
       p*=x;
   }
   cout<<"\n"<<s;
   return 0;
} Natija :

Labaratoriya №2 Mavzu: C++ tilida takrorlanuvchi jarayonlar dasturlar tuzish. For,while va do while takrorlash operatorlari. Misollar

1. Misol. (Namuna) Butun son berilgan agar a>0 bo’lsa a+1 aks holsa a hisoblansin. #include <iostream.h> #include <math.h> int main () { int a; cin>>a; if(a>0){a++;} cout<<a; return 0; } Javob: 1-misol. Dastur kodi: #include <iostream> using namespace std; int main() { int n; double s=0; cin>>n; for(int i=1; i<=n; i++) { s+=1./i; } cout<<s; return 0; }

Natija: 2-misol. Dastur kodi: #include <iostream> using namespace std; int main() { int n, s=0; cin>>n; for(int i=n; i<=2*n; i++) { s+=i*i; } cout<<s; return 0; }

Natija: 3-misol. Dastur kodi: #include <iostream> using namespace std; int main() { int n; double p=1,k=1; cin>>n; for(int i=1; i<=n; i++) { k+=0.1; p*=k; } cout<<p; return 0; }

Natija : 4-misol. Dastur kodi: #include <iostream> using namespace std; int main() { double A, p; int n; cin>>A; cin>>n; p=A; for(int i=1; i<=n; i++) { cout<<p<<" "; p*=A; } return 0; }